Maxwell's Equations and the Variation of Metric Determinant

[etotheipi]

Homework Statement

Derive the source-free Maxwell equations from the action$$S = - \frac{1}{2} \int F \wedge \star F$$

Relevant Equations

Some wedgey things

First, I let $\omega = \sqrt{-g} dx^0 \wedge \dots \wedge dx^3$ be the top-form on $M$, and making use of the inner-product on the space of forms I can write$$\begin{align*}

F \wedge \star F = g(F,F) \omega &= \frac{1}{4} F_{\mu \nu} F_{\rho \sigma}
\begin{vmatrix}
g(dx^{\mu}, dx^{\rho}) & g(dx^{\mu}, dx^{\sigma})\\
g(dx^{\nu}, dx^{\rho}) & g(dx^{\nu}, dx^{\sigma})
\end{vmatrix} \omega \\

&= \frac{1}{4} F_{\mu \nu} F_{\rho \sigma} (\delta^{\mu \rho} \delta^{\nu \sigma} - \delta^{\nu \rho} \delta^{\mu \sigma}) \omega\\

&= \frac{1}{4}\left( F^{\rho \sigma} F_{\rho \sigma} - F^{\sigma \rho} F_{\rho \sigma} \right) \omega = \frac{1}{2} F^{\sigma \rho} F_{\rho \sigma} \omega

\end{align*}$$due to the anti-symmetry of $F^{\rho \sigma}$. Replacing $\omega$ gives $F \wedge \star F = \frac{1}{2} F^{\rho \sigma} F_{\rho \sigma} \sqrt{-g} d^4 x$, where I abbreviated $d^4 x \equiv dx^0 \wedge \dots \wedge dx^3$. Now we can vary the action$$\delta S = - \frac{1}{4} \int d^4 x \left(\delta(\sqrt{-g}) F^{\rho \sigma} F_{\rho \sigma} + \sqrt{-g} F^{\rho \sigma} \delta(F_{\rho \sigma}) + \sqrt{-g} \delta(F^{\rho \sigma}) F_{\rho \sigma}\right)$$Because of following property of diagonal matrices $A$,$$\frac{1}{\mathrm{det} A} \delta(\mathrm{det} A) = \mathrm{tr}(A^{-1} \delta A)$$we can write down the variation of the metric determinant$$\delta(\sqrt{-g}) = - \frac{1}{2} \sqrt{-g} g_{\mu \nu} \delta(g^{\mu \nu})$$But I don't know how to re-write the variations of the $F_{\rho \sigma}$ and $F^{\rho \sigma}$. Wondered if someone could give a few pointers? Thanks

(Also, whilst I'm here, is there a faster/better way to work out $F \wedge \star F$?)

 

[atyy]

Not sure if this is helpful: http://davinci.fmph.uniba.sk/~fecko1/referaty/regensburg_2007.pdf (p21)

 

[etotheipi]

Not sure if this is helpful: http://davinci.fmph.uniba.sk/~fecko1/referaty/regensburg_2007.pdf (p21)

Ooh, those look quite nice! Give me like, half an hour or so to go through them. The problem is that right now I don't really have an intuitive feel for exterior calculus (I can't feel it in my bones yet..., you know ), so I'm really uncertain about many of these manipulations.

 

[atyy]

If you have access via a library to his book https://www.amazon.com/dp/0521187966/?tag=pfamazon01-20, there is also some discussion in section 16.3.2.

 

[samalkhaiat]

Homework Statement:: Derive the source-free Maxwell equations from the action$$S = - \frac{1}{2} \int F \wedge \star F$$
Relevant Equations:: Some wedgey things
F \wedge \star F = \frac{1}{2} F^{\sigma \rho} F_{\sigma \rho} \omega .
Now we can vary the action
\delta S = - \frac{1}{4} \int d^4 x \left(\delta(\sqrt{-g}) F^{\rho \sigma} F_{\rho \sigma} + \sqrt{-g} F^{\rho \sigma} \delta(F_{\rho \sigma}) + \sqrt{-g} \delta(F^{\rho \sigma}) F_{\rho \sigma}\right)
But I don't know how to re-write the variations of the $F_{\rho \sigma}$ and $F^{\rho \sigma}$. Wondered if someone could give a few pointers? Thanks
(Also, whilst I'm here, is there a faster/better way to work out $F \wedge \star F$?)

1) To obtain the equations of motion, you need to vary the action with respect to the dynamical variables.

2) Varying the action with respect to the background metric gives you the energy-momentum tensor.

3) If \alpha , \beta \in \Lambda^{p}(M^{n}), one can show (with some work) that \alpha \wedge \ast \beta = \beta \wedge \ast \alpha = (\alpha , \beta ) \ \mu , \ \ \ \ \ (1) where (\alpha , \beta)(x) = \frac{1}{p!} \alpha_{\mu_{1} \cdots \mu_{p}}(x) \ \beta^{\mu_{1} \cdots \mu_{p}}(x), \mu = \frac{\sqrt{-g}}{n!} \ \epsilon_{\mu_{1} \cdots \mu_{n}} \ dx^{\mu_{1}} \wedge \cdots \wedge dx^{\mu_{n}} = \sqrt{-g} \ d^{n}x . So for F \in \Lambda^{2}(M^{1,3}), the Hodge star of F is given by \ast F = \frac{1}{2} \mathscr{F}_{\mu\nu} \ dx^{\mu} \wedge dx^{\nu} , where \mathscr{F}_{\mu\nu} = \frac{\sqrt{-g}}{2} \ \epsilon_{\mu\nu\rho\sigma} F^{\rho\sigma}. Thus, from (1), you get F \wedge \ast F = (F , F) \ \mu = \frac{\sqrt{-g}}{2} F_{\mu\nu} \ F^{\mu\nu} \ d^{4}x . \ \ \ \ \ (2)

4) Without using (2), Maxwell’s equations can be derived directly from the action S[A] = - \int_{\Omega} \left( \frac{1}{2}F \wedge (\ast F) + A \wedge (\ast j)\right) , where \ast j = \frac{1}{3!} j^{\mu} \ \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma} is the 3-form associated with the 4-vector current j^{\mu}. Under A \to A + \epsilon \alpha, with \epsilon \ll 1 and \alpha|_{\partial \Omega} = 0 but otherwise arbitrary 1-form, we have (using F = dA) S[A + \epsilon \alpha ] = S[A] - \epsilon \left( \int_{\Omega} \left( \frac{1}{2} F \wedge \ast d\alpha + \frac{1}{2} d\alpha \wedge \ast F + \alpha \wedge \ast j \right) \right) + \mathcal{O}(\epsilon^{2}) . Use (1) to obtain \delta S = \lim_{\epsilon \to 0} \frac{S[A + \epsilon \alpha] - S[A]}{\epsilon} = - \int_{\Omega} \left( d\alpha \wedge \ast F + \alpha \wedge \ast j \right). Now, integrate the first term by parts using d\left( \alpha \wedge \ast F \right) = d\alpha \wedge \ast F - \alpha \wedge d \ast F, then use Stokes’ theorem: \delta S[A] = - \int_{\partial \Omega} \alpha \wedge \ast F - \int_{\Omega} \alpha \wedge \left( d \ast F + \ast j \right) . The first integral vanish because \alpha |_{\partial \Omega} = 0. Thus, the action is stationary if d \ast F + \ast j = 0. \ \ \ \ (3) Since j is 1-form on the 4-dimensional Minkowski space, then \ast^{2}j = j. Therefore (3) may also be written in the form \ast d \ast F = \delta F = - j , \ \ \ \ \ (4) where \delta : \Lambda^{p}(M^{n}) \to \Lambda^{p - 1}(M^{n}) is the so-called codifferential operator.

Here is an exercise for you: Show that (3) [or (4)] reproduces the Maxwell equation \partial_{\mu}F^{\mu\nu} = j^{\nu} .

 

[vanhees71]

I start to love the good old Ricci calculus more and more ;-)). SCNR.

 

[etotheipi]

Thanks, that's great! Okay, for the final part, I found it a little bit tricky to keep track of all of the $\sqrt{-g}$'s..., but I'll show what I tried in any case. Using equation (2.27) from Tong's GR notes, $(d\omega)_{\mu_1 \dots \mu_{p+1}} = (p+1) \partial_{[ \mu_1} \omega_{\mu_2 \dots, \mu_{p+1}]}$, as well as the identity $\epsilon_{abcd} \epsilon^{defg} = 6\delta_{[a}^e \delta_{b}^{f} \delta_{c]}^{g}$, we obtain$$\begin{align*}

(\star j)_{\nu \rho \sigma} = \sqrt{-g} \epsilon_{\mu \nu \rho \sigma} j^{\mu} = - (d \star F)_{\nu \rho \sigma} = -3 \partial_{[\nu} (\star F)_{\rho \sigma]} &= - \frac{(-g)}{2} \epsilon_{\nu \rho \sigma \mu}\, \epsilon^{\mu \alpha \beta \gamma} \partial_{\alpha} (\star F)_{\beta \gamma} \\

&= (-g) \epsilon_{\nu \rho \sigma \mu} \partial_{\alpha} \left(- \frac{1}{2} \epsilon^{\mu \alpha \beta \gamma} (\star F)_{\beta \gamma}\right)

\end{align*}$$Now since $\star(\star F) = \pm F$, taking the Hodge dual of the definition of $\star F$, and then using the metric to raise the free indices and also to raise/lower both pairs of dummy indices, gives$$(\star F)_{\beta \gamma} = \frac{\sqrt{-g}}{2} \epsilon_{\beta \gamma \delta \varepsilon} F^{\delta \varepsilon} \implies F^{\beta \gamma} = \frac{\sqrt{-g}}{2} \epsilon^{\beta \gamma \delta \varepsilon} (\star F)_{\delta \varepsilon}
$$In other words, the bit inside the brackets can be re-written as$$ -\frac{1}{2} \epsilon^{\mu \alpha \beta \gamma} (\star F)_{\beta \gamma} = \frac{-1}{\sqrt{-g}} F^{\mu \alpha} = \frac{1}{\sqrt{-g}} F^{\alpha \mu}$$So the first equation becomes, after cycling each of the indices on the Levi-Civita symbol on the RHS one place to the right, and cancelling the $\sqrt{-g}$ terms on either side,$$\epsilon_{\mu \nu \rho \sigma} j^{\mu} = \epsilon_{\mu \nu \rho \sigma} \partial_{\alpha} F^{\alpha \mu} \implies j^{\mu} = \partial_{\alpha} F^{\alpha \mu}$$How does that look?

 

[samalkhaiat]

I found it a little bit tricky to keep track of all of the $\sqrt{-g}$'s

The exercise asks you to reproduce Maxwell’s equation on the 4-dimensional Minkowski space. So, you can set \sqrt{-g} = 1. If you keep \sqrt{-g(x)}, then Eq(3) or (4) reproduce the corresponding Maxwell’s equation on the 4-dimensional pseudo-Riemannian space \nabla_{\mu}F^{\mu\nu} = j^{\nu}, as explained bellow.

How does that look?

I did not check it because it looks very messy.

I will choose to derive \nabla_{\mu}F^{\mu\nu} = j^{\nu} from Eq(4) because it is lengthier and require repeated application of the \ast operator. This will let you learn how to keep track of the many metric tensors and the \sqrt{-g} factors. Let’s start from the basic facts. We have, for F \in \Lambda^{2}(M^{(1,3)}),

F = \frac{1}{2!} F_{\mu\nu} \ dx^{\mu} \wedge dx^{\nu} . Thus,

\ast F = \frac{1}{2} F_{\mu\nu} \ \ast (dx^{\mu} \wedge dx^{\nu}) .

Since \ast : \Lambda^{p}(M^{n}) \to \Lambda^{n - p}(M^{n}), then the action of \ast on the element dx^{\mu} \wedge dx^{\nu} of the local basis of \Lambda^{2}(M^{(1,3)}) is given by

\ast (dx^{\mu} \wedge dx^{\nu}) = \frac{\sqrt{-g}}{(4 - 2)!} \ g^{\mu\rho}g^{\nu\sigma} \ \epsilon_{\rho\sigma\alpha\beta} \ dx^{\alpha} \wedge dx^{\beta} .

Thus \ast F = \frac{\sqrt{-g}}{4} F^{\rho\sigma} \ \epsilon_{\rho\sigma\alpha\beta} \ dx^{\alpha} \wedge dx^{\beta}. Taking the exterior derivative d : \Lambda^{p}(M^{n}) \to \Lambda^{p + 1}(M^{n}), we get

d \ast F = \frac{1}{4} \partial_{\tau}(\sqrt{-g}F^{\rho\sigma}) \ \epsilon_{\rho\sigma\alpha\beta} \ dx^{\tau} \wedge dx^{\alpha} \wedge dx^{\beta} . Operating with \ast: \ast d \ast F = \frac{1}{4} \partial_{\tau}(\sqrt{-g}F^{\mu\nu}) \ \epsilon_{\mu\nu\alpha\beta} \ \ast (dx^{\tau} \wedge dx^{\alpha} \wedge dx^{\beta}) . \ \ \ (1) Now

\ast (dx^{\tau} \wedge dx^{\alpha} \wedge dx^{\beta}) = \frac{\sqrt{-g}}{(4 - 3)!} \ g^{\tau \mu} g^{\alpha \nu} g^{\beta \gamma} \ \epsilon_{\mu \nu \gamma \eta} \ dx^{\eta} . \ \ \ (2) From the definition of the determinate of the inverse metric: \epsilon_{\mu \nu \gamma \eta} \ g^{\mu \tau} g^{\nu \alpha} g^{\gamma \beta} g^{\eta \delta} = \frac{1}{g} \ \epsilon^{\delta \tau \alpha \beta} , it follows that \epsilon_{\mu \nu \gamma \eta} \ g^{\mu\tau} g^{\nu \alpha} g^{\gamma \beta} = \frac{1}{g} \ \epsilon^{\rho \tau \alpha \beta} \ g_{\rho \eta} . Substituting in (2), we get \ast (dx^{\tau} \wedge dx^{\alpha} \wedge dx^{\beta}) = - \frac{1}{\sqrt{-g}} \ \epsilon^{\rho \tau \alpha \beta} \ g_{\rho \eta} \ dx^{\eta} . Substituting in (1) leads to \ast d \ast F = \frac{-1}{4 \sqrt{-g}} \partial_{\tau}(\sqrt{-g}F^{\mu\nu}) \ \epsilon_{\mu \nu \alpha \beta} \ \epsilon^{\rho \tau \alpha \beta} \ g_{\rho \eta} \ dx^{\eta} . Finally, we make use of the identity \epsilon_{\mu \nu \alpha \beta} \ \epsilon^{\rho \tau \alpha \beta} \ g_{\rho \eta} = (-2!) \left( \delta^{\tau}_{\nu} \ g_{\mu\eta} - \delta^{\tau}_{\mu} \ g_{\nu\eta} \right), to obtain \ast d \ast F = - \left[ \frac{1}{\sqrt{-g}} \ \partial_{\nu} \left(\sqrt{-g} F^{\nu\mu}\right)\right] \ g_{\mu\eta} \ dx^{\eta} . Since the tensor F^{\mu\nu} is antisymmetric, the expression in the big brackets is nothing but the (spacetime) covariant expression \nabla_{\nu}F^{\nu\mu}. So, we are almost done: \ast d \ast F = - \nabla_{\nu}F^{\nu\mu} \ g_{\mu\eta} \ dx^{\eta} . \ \ \ \ \ \ \ (3) Now, the 1-form associated with the current vector is given by j = j^{\mu} \ g_{\mu \eta} \ dx^{\eta} . \ \ \ \ \ \ \ \ \ \ \ \ \ (4) So, from \ast d \ast F + j = 0, it follows that \nabla_{\nu}F^{\nu\mu} = j^{\mu} .

Exercise: Start with the 3-form \ast j = \frac{\sqrt{-g}}{3!} \ j^{\mu} \ \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma} . Show that operating with \ast gives you j = j^{\mu} \ g_{\mu \eta} \ dx^{\eta} .

 

[etotheipi]

Exercise: Start with the 3-form \ast j = \frac{\sqrt{-g}}{3!} \ j^{\mu} \ \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma} . Show that operating with \ast gives you j = j^{\mu} \ g_{\mu \eta} \ dx^{\eta} .

Thanks! I'll have a go at this exercise. Since $*j \in \Lambda^3(M^{(3,1)})$, it satisfies $*(*j) = j$. Let's also make use of the equation$$*(dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma}) = \sqrt{-g} g^{\nu \alpha} g^{\rho \beta} g^{\sigma \gamma} \epsilon_{\alpha \beta \gamma \eta} dx^{\eta}$$in order to write down$$\begin{align*}
j = *(*j) &= \frac{\sqrt{-g}}{3!} j^{\mu} \epsilon_{\mu \nu \rho \sigma} *(dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma}) \\ \\

j &= \frac{(-g)}{3!} j^{\mu} \epsilon_{\mu \nu \rho \sigma} g^{\nu \alpha} g^{\rho \beta} g^{\sigma \gamma} \epsilon_{\alpha \beta \gamma \eta} dx^{\eta} \ \ \ (1)

\end{align*}$$Now since$$g^{\nu \alpha} g^{\rho \beta} g^{\sigma \gamma} \epsilon_{\alpha \beta \gamma \eta} = \frac{1}{g} \epsilon^{\lambda \nu \rho \sigma} g_{\lambda \eta}$$we can re-write $(1)$ as$$j = \frac{(-g)}{3!} j^{\mu} \epsilon_{\mu \nu \rho \sigma} \left( \frac{1}{g} \epsilon^{\lambda \nu \rho \sigma} g_{\lambda \eta} \right) dx^{\eta} = \frac{(-1)}{3!} j^{\mu} \epsilon_{\mu \nu \rho \sigma} \epsilon^{\lambda \nu \rho \sigma} g_{\lambda \eta} dx^{\eta} \ \ \ (2)$$Then, finally we can use the identity$$\epsilon_{\mu \nu \rho \sigma} \epsilon^{\lambda \nu \rho \sigma} = (3!) \delta^{\lambda}_{\mu}$$to re-write (2) as$$j = \frac{(-1)}{3!} j^{\mu} (3!) \delta^{\lambda}_{\mu} g_{\lambda \eta} dx^{\eta} = - j^{\mu} g_{\mu \eta} dx^{\eta}$$I seem to have picked up an erroneous negative sign somewhere... ...

 

[samalkhaiat]

finally we can use the identity \epsilon_{\mu \nu \rho \sigma} \epsilon^{\lambda \nu \rho \sigma} = (3!) \delta^{\lambda}_{\mu}
I seem to have picked up an erroneous negative sign somewhere... ...

\epsilon_{\mu \nu \rho \sigma} \ \epsilon^{\lambda \nu \rho \sigma} = (-3!) \ \delta^{\lambda}_{\mu}
so that \epsilon^{\mu\nu\rho\sigma} \ \epsilon_{\mu\nu\rho\sigma} = -4!, with \epsilon^{0123} = +1.

 

[etotheipi]

\epsilon_{\mu \nu \rho \sigma} \ \epsilon^{\lambda \nu \rho \sigma} = (-3!) \ \delta^{\lambda}_{\mu}
so that \epsilon^{\mu\nu\rho\sigma} \ \epsilon_{\mu\nu\rho\sigma} = -4!, with \epsilon^{0123} = +1.

Ah! Careless of me. Thank you